Andrew Ng的machine learning课程week2编程题python实现

# -*- coding:utf-8 -*-import numpy as npimport matplotlib.pylab as pltfrom multiprocessing import Processdef plotData(X, y):    plt.scatter(X, y, color='r', s=10, marker='x')    plt.xlabel('Population of City in 10,000s')    plt.ylabel('Profit in $10,000s')    plt.show()def computerCost(X, y, theta):    m = len(y)    h = np.dot(X, theta)  #X和theta是array类型，用点乘    square = np.square(h - y)     J = 1/(2*m)*sum(square)    return Jdef gradientDescent(X, y, theta, alpha, num_iters):    m = len(y)    J_history = np.zeros((num_iters,1))    for i in range(num_iters):        theta = theta - alpha/m*(np.dot(X.T,(np.dot(X,theta))-y))        J_history = computerCost(X, y, theta)    return theta, J_historyif __name__ == '__main__':    #X为txt里第一列    X = np.loadtxt(fname='E:\python\ex1_multi\data\ex1data1.txt',dtype=float,delimiter=",",usecols=(0,))    #y为txt里第二列    y = np.loadtxt(fname='E:\python\ex1_multi\data\ex1data1.txt',dtype=float,delimiter=",",usecols=(1,))    m = len(X)    #读取的数据为1*m,转换为m*1    X = X.reshape(m, 1)    y = y.reshape(m, 1)    #绘图    plotData(X, y)    X = np.column_stack((np.ones((m,1),dtype=int), X)) #Add a column of ones to x    theta = np.zeros((2,1))    iterations = 1500    alpha = 0.01    J = computerCost(X, y, theta) #Expected cost value (approx) 32.07    print(J)    J = computerCost(X, y, [[-1],[2]]) #Expected cost value (approx) 54.24    print(J)    theta = gradientDescent(X, y, theta, alpha, iterations)[0]    print(theta)